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The Heisenberg uncertainty principle for position and impulse gives for the uncertainty in position $\Delta x$ and uncertainty in impulse $\Delta p$ that $ \Delta x \cdot \Delta p \geq \frac{1}{2}\hbar$.

If a particle that is not moving with the speed of light does not have a mass, then we always have $m=0$ and hence $p=0$ with certainty, and hence $\Delta p = 0$. If the particle is moving with the speed of light, I think we can have $p>0$ and also $\Delta p > 0$, using relativity theory.

However, it is not possible that $\Delta p = 0$, because it gives $\Delta x \cdot \Delta p =0$, contradicting the Heisenberg uncertainty principle. Therefore, I think this implies one of the following things:

  • It is not possible to know for certain that $m=0$. This is difficult; we know that light always moving with the speed of light in vacuum (that is one of the axioms of relativity), and that is only possible if $m=0$ by relativity.
  • It is not possible to know for a massless particle whether $v<c$ or $v=c$. This doesn't give a direct contradiction. Is this the case?

I might be confusing several things form several areas and apply them in the wrong way, so I'd like to know whether this makes sense or not, and if it doesn't then what the Heisenberg uncertainty principle means for massless particles.

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    $\begingroup$ "If a particle that is not moving with the speed of light does not have a mass"...massless particles always move with the speed of light, and the HUP is entirely unrelated to this. $\endgroup$ – ACuriousMind Dec 8 '16 at 17:09
  • $\begingroup$ @ACuriousMind Why? Just because no force is needed to accelarate them? (In that case, there is also no force is needed to decelarate them, so that is not really convincing) I am largely self-studying these areas of physics, so I might have "missed" some things that I should know while studying this. $\endgroup$ – wythagoras Dec 8 '16 at 17:12
  • $\begingroup$ see physics.stackexchange.com/q/90469/50583 $\endgroup$ – ACuriousMind Dec 8 '16 at 17:13
  • $\begingroup$ The fact that massless particles always move at the same fixed velocity c means that they are always totally delocalized in the whole 3D space, in other words they must be a plane wave. When absorbed, e.g., as a photon, they are completely localized but by then they do not move any more. $\endgroup$ – hyportnex Dec 8 '16 at 18:52
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Massless particles always move at the speed of light.

For massless particles especially, it is not true that the momentum: $p = mv$. In general, momentum is actually calculated as, $$(pc)^2 = E^2 - (mc^2)^2$$ sometimes called the "relativistic momentum". For low velocities (and non-zero masses) this can be approximated as $p \approx mv$.

This means that for massless particles, $p = E/c$, and the energy can never be (identically) zero---so there's never an issue with the uncertainty principle. Also, keep in mind that even if $p=0$, that doesn't necessarily mean that $\Delta p = 0$.

Final, minor (and semantic note): "impulse" refers to a specific change in momentum of a body, or the action that causes such a change in momentum. $\Delta p$ isn't really an 'impulse' per se in the context of the HUP; instead it's a "momentum uncertainty" or "momentum localization" etc.

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